(D^2-4D+4)y=100

Solution for (D^2-4D+4)y=100 equation:

(^2-4+4)D=100

We move all terms to the left:

(^2-4+4)D-(100)=0

We multiply parentheses

D^2-4D+4D-100=0

We add all the numbers together, and all the variables

D^2-100=0

a = 1; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·1·(-100)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*1}=\frac{-20}{2}
=-10
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*1}=\frac{20}{2}
=10
$